map＜int,float＞ mp+mp.insert(make_pair(x,y))的使用，map的遍历必须使用迭代器实现

1009 Product of Polynomials (25分)

This time, you are supposed to find A×B where A and B are two polynomials.

Input Specification:

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:

K N​1​​ a​N​1​​​​ N​2​​ a​N​2​​​​ ... N​K​​ a​N​K​​​​

where K is the number of nonzero terms in the polynomial, N​i​​ and a​N​i​​​​ (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10, 0≤N​K​​<⋯<N​2​​<N​1​​≤1000.

Output Specification:

For each test case you should output the product of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate up to 1 decimal place.

Sample Input:

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output:

3 3 3.6 2 6.0 1 1.6

AC代码：

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<string.h>
#include<string>
#include<stack>
#include<cmath>
#include<map>
using namespace std;
#pragma warning(disable:4996)
const int maxn = 10000 + 10;
float ans[maxn];
map<int, float> mp1;
map<int, float> mp2;
int main(){
	memset(ans, 0, sizeof(ans));
	int k1, k2,e,e1,e2;
	float c,c1,c2;
	scanf("%d", &k1);
	for (int i = 0; i < k1; i++) {
		scanf("%d %f", &e, &c);
		mp1.insert(make_pair(e, c));
	}
	scanf("%d", &k2);
	for (int i = 0; i < k2; i++) {
		scanf("%d %f", &e, &c);
		mp2.insert(make_pair(e, c));
	}
	for (map<int, float> ::iterator it1 = mp1.begin(); it1 != mp1.end(); it1++) {
		e1 = it1->first;
		c1 = it1->second;
		for (map<int, float> ::iterator it2 = mp2.begin(); it2 != mp2.end(); it2++) {
			e2 = it2->first;
			c2 = it2->second;
			ans[e1 + e2] += c1 * c2;
		}
	}
	int countn = 0;
	for (int i = 0; i < maxn; i++) {
		if (fabs(ans[i] - 0.0) > 1e-6)
			countn++;
	}
	printf("%d", countn++);
	for (int i = maxn; i >= 0;i--) {
		if (fabs(ans[i] - 0.0) > 1e-6)
			printf(" %d %.1f", i,ans[i]);
	}
	return 0;
}